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Small-argument cancellation in the upward Boys recurrence

Example

For $0 < T \leq \frac{1}{4}$ , the upward formula for the first two Boys orders is $F_{1} (T) = \frac{F_{0} (T) - e^{- T}}{2 T} .$ Both subtracted quantities are close to one, while their difference is of order $T$ . If the input intervals for them have radii $r_{0}, r_{e}$ , the exact upward affine image has radius $(r_{0} + r_{e}) / (2 T)$ . If its endpoints are converted outward once with finite errors $δ_{-}, δ_{+}$ , the stored radius is $(r_{0} + r_{e}) / (2 T) + (δ_{-} + δ_{+}) / 2$ . Rounding at intermediate nodes requires propagating their contributions as well. Thus fixed input radii can be strongly amplified as $T ↓ 0$ . The direct Taylor branch has the certified remainder $T^{2} / 10$ for $F_{0}$ after its linear term, so it avoids this division by a small $T$ .

Facts & Assumptions

Given: $0 < T \leq 1 / 4$ , the Boys integral and recurrence, and finite certified intervals for $F_{0} (T)$ and $e^{- T}$ .

[F1]

Boys integral, power series, recurrence, and incomplete-gamma representation: $F_{1} = (F_{0} - e^{- T}) / (2 T)$ and $F_{0} (T) = 1 - T / 3 + \dots$ .

[F2]

The real exponential function and the number $e$ by a power series: $e^{- T} = \sum_{k = 0}^{\infty} (- T)^{k} / k!$ .

[F3]

A monotone-range Taylor remainder for small Boys arguments: after $1 - T / 3$ , the $F_{0}$ remainder has absolute value at most $T^{2} / 10$ for $T \leq 1$ .

[F4]

Upward and downward Boys recurrence error amplification bounds: the exact upward image has radius $(r_{0} + r_{e}) / (2 T)$ ; one finite outward conversion of each full-image endpoint adds the mean endpoint error. Intermediate-node errors must instead be propagated through all subsequent affine coefficients.

[F5]

Soundness of every successful finite-budget Boys enclosure and Finite-budget certified evaluation of Boys function tables: the evaluator offers a small- $T$ Taylor enclosure and makes no success claim when a finite recurrence candidate misses tolerance.

Verification

1.1givenF1F2F3algebra

The exact identity in [F1] divides a difference of two terms. By [F1] and [F3], $F_{0} (T) = 1 - T / 3 + R_{F}$ with $∣ R_{F} ∣ \leq T^{2} / 10$ . By [F2], $e^{- T} = 1 - T + R_{e}$ with $∣ R_{e} ∣ \leq T^{2}$ for $T \leq 1$ : for $k \geq 2$ , $k! \geq 2^{k - 1}$ , so $\sum_{k \geq 2} 1 / k! \leq \sum_{k \geq 2} 2^{1 - k} = 1$ . Therefore $∣ (F_{0} (T) - e^{- T}) - \frac{2 T}{3} ∣ \leq \frac{11}{10} T^{2},$ so for $T \leq 1 / 4$ the positive difference is at least $T / 4$ . The two terms being subtracted each tend to one as $T ↓ 0$ .

2.1givenF3F4F5algebra∎

Applying [F4] with $m = 0$ gives a factor $1 / (2 T)$ on both seed radii, which diverges as $T ↓ 0$ for fixed nonzero radii. The stated stored-radius formula follows only for a full affine image with one finite conversion per endpoint; a primitive subtraction error must also be divided by $2 T$ before adding the final division error. In contrast, [F3] gives a direct Taylor enclosure with error $T^{2} / 10$ . The algorithm [F5] can therefore select the direct small-argument branch and return only a certified interval; the example makes no claim that every requested tolerance succeeds under every finite cap.

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Sources

Trygve Helgaker, Molecular Integral Evaluation, 10th Sostrup Summer School lecture notes (2008) (standard reference, not scraped)
L. E. McMurchie and E. R. Davidson, One and Two Electron Integrals over Cartesian Gaussian Functions, UCRL-13791 (1976) (standard reference, not scraped)